After introducing the properties of numbers, Spivak states:

"Note, in particular, that $a>0$ if and only if $a$ is in $P$"

I'm not exactly sure how to prove this. The relevant properties are:

"(P10) (Trichotomy law) For every number $a$, one and only one of the following holds:
(i) $a = 0$
(ii) $a$ is in the collection $P$
(iii) $-a$ is in the collection $P$"

and the definition:

"$a>b$ if $a-b$ is in $P$"

Using a previously introduced property and the definition:

$a$ is in $P \implies a>0$

But how do I show:

$a>0 \implies a$ is in $P$?

  • $\begingroup$ we don't all have the book in front of us. What is P? and how was it defined. It sounds like $a > 0$ if and only if $a \in P$ sounds like a definition? Are is there something like you can add something to it to get zero? Or what. $\endgroup$ – fleablood Feb 16 at 0:12

The definition of $a \gt 0$ tells you that $a-0 \in P$, and $a-0=a$ so $a-0 \in P \Rightarrow a \in P$.

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    $\begingroup$ If A then B, does not imply if B then A? $\endgroup$ – user1390010 Feb 15 at 23:09
  • $\begingroup$ @user1390010 Definitions are if and only if statements, always. $\endgroup$ – Pedro Tamaroff Feb 15 at 23:10
  • $\begingroup$ That's a good question, but in the context of a definition, @PedroTamaroff is correct. $\endgroup$ – Robert Shore Feb 15 at 23:13
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    $\begingroup$ OK thanks, just one more clarification: does "A if B" always mean "A if and only if B" or only in the context of a definition? (since it is not in the usual "if B then A" structure) $\endgroup$ – user1390010 Feb 15 at 23:17
  • $\begingroup$ It really depends on context, but you can pretty much count on that being the case in the context of a definition (because otherwise the language isn't actually defining the concept). $\endgroup$ – Robert Shore Feb 15 at 23:29

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